Sel*_*man 3 python cookies selenium pickle webdriverwait
我正在尝试制作一个 selenium instagram 脚本。我的问题是我每次都必须登录我的帐户。
\n如果我使用普通的谷歌浏览器,只需登录一次就足够了。当 \xc4\xb1 打开 Chrome 时,它始终保持登录状态。但 Chrome 驱动程序总是要求登录。
\nfrom selenium import webdriver\nimport time\nimport random\nfrom selenium.webdriver.common.by import By\nimport kullaniciBilgileri as kb\nimport os.path\nfrom pathlib import Path\nfrom selenium.webdriver.chrome.options import Options\nimport collections\n\noptions = Options()\noptions.add_experimental_option('useAutomationExtension', False)\nlink = "http://www.instagram.org"\nbrowser = webdriver.Chrome(options=options, executable_path="./chromedriver")\nbrowser.get(url=link)\ntime.sleep(10)\nusername = browser.find_element(by=By.NAME, value=("username"))\npassword = browser.find_element(by=By.NAME, value=("password"))\nusername.send_keys(kb.userName)\npassword.send_keys(kb.password)\nloginBtn = browser.find_element(by=By.CSS_SELECTOR, value=("#loginForm > div > div:nth-child(3) > button > div"))\n \nloginBtn.click()\nRun Code Online (Sandbox Code Playgroud)\n有什么建议吗?
\n在访问Instagram以避免每次登录时,一旦您第一次使用pickle模块登录,您可以存储cookie并在下次登录尝试时重复使用,如下所示:
\nfrom selenium.webdriver.support.ui import WebDriverWait\nfrom selenium.webdriver.common.by import By\nfrom selenium.webdriver.support import expected_conditions as EC\nimport pickle\n\ndriver.execute("get", {\'url\': \'http://www.instagram.org\'})\nWebDriverWait(driver, 20).until(EC.element_to_be_clickable((By.CSS_SELECTOR, "input[name=\'username\']"))).send_keys("_SelmanFarukY\xc4\xb1lmaz_")\ndriver.find_element(By.CSS_SELECTOR, "input[name=\'password\']").send_keys("Selman_Faruk_Y\xc4\xb1lmaz")\ndriver.find_element(By.CSS_SELECTOR, "button[type=\'submit\'] div").click()\npickle.dump( driver.get_cookies() , open("cookies.pkl","wb"))\ndriver.quit()\ndriver = webdriver.Chrome(service=s, options=options)\ndriver.execute("get", {\'url\': \'http://www.instagram.org\'})\ncookies = pickle.load(open("cookies.pkl", "rb"))\nfor cookie in cookies:\n driver.add_cookie(cookie)\ndriver.execute("get", {\'url\': \'http://www.instagram.org\'})\nRun Code Online (Sandbox Code Playgroud)\n
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